Integrate dx/[x + sqrt (1 + x^2)]^n?

Make the Euler substitution:
Let x= (u^2-1) /(2u)
x = (1/2) u^(-1) (u^2-1)
dx = ( (1/2) (-1)u^(-2) (u^2-1) + (1/2) u^(-1) (2u) ) du
dx = (- (u^2-1) /(2u^2) + 1) du
dx = ( 1 - (u^2-1) /(2u^2) ) du


x = (u^2-1) /(2u)
2ux = u^2-1
u^2 = 2ux+1
substitute x= (u^2-1) /(2u)

x+ sqrt(1+x^2) = (u^2-1) /(2u) + sqrt ( 1+ (u^2-1)^2 / (4u^2) )
x+ sqrt(1+x^2) = (u^2-1) /(2u) + sqrt ( 4u^2+ (u^2-1)^2) /(2u)
x+ sqrt(1+x^2) = [(u^2-1) + sqrt ( 4u^2+ (u^2-1)^2)] /(2u)
x+ sqrt(1+x^2) = [(u^2-1) + sqrt ( 4u^2+ u^4-2u^2+1] /(2u)
x+ sqrt(1+x^2) = [(u^2-1) + sqrt ( u^4+2u^2+1] /(2u)
x+ sqrt(1+x^2) = [(u^2-1) + sqrt ( (u^2+1)^2] /(2u)
x+ sqrt(1+x^2) = [(u^2-1) + (u^2+1)] /(2u)
x+ sqrt(1+x^2) = [2u^2] /(2u)
x+ sqrt(1+x^2) = u
u = x+ sqrt(1+x^2)

∫ dx/[x + sqrt (1 + x^2)]^n = ∫ u^(-n) ( 1 - (u^2-1) /(2u^2) ) du
∫ dx/[x + sqrt (1 + x^2)]^n = ∫ u^(-n) ( 2u^2 - (u^2-1) /(2u^2) ) du
∫ dx/[x + sqrt (1 + x^2)]^n = (1/2) ∫ u^(-n-2) ( u^2 +1) du
= (1/2) ∫ u^(-n) du + (1/2) ∫ u^(-n-2) du
= (1/2) u^(-n+1) /(-n+1) + (1/2) u^(-n-2+1) /(-n-2+1)
= (1/2) u^(1-n) /(1-n) + (1/2) u^(-n-1) /(-n-1)

replace u by x+ sqrt(1+x^2)
= (1/2) (x+sqrt(1+x^2))^(1-n) /(1-n) + (1/2) (x+sqrt(x^2+1))^(n--1) / (-n-1) + C

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Integral dx/[x + sqrt (1 + x^2)]^n

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Set x = (t² - 1)/2t....(*)
dx = (1/2 + 1/2t²)dt
x + √(1 + x²) = (t² - 1)/2t + √(1 + (t⁴ - 2t² + 1)/4t²) = (t² - 1)/2t + (t² + 1)/2t = t
∫dx/[x + √(1 + x²)]⁻ⁿdx = ∫(t⁻ⁿ/2 + t⁻ⁿ ⁻ ²/2)dt = (1/2)(t¹⁻ⁿ/(1 - n) + (1/2)t⁻³⁻ⁿ/(-3 - n). Solving (*), t = x + √(1 + x²). So the integral is {1/2(1 -n)}[x + √(1 + x²)]^(1 - n) - {1/2(1 + n)}[x + √(1 + x²)]^ (-n -3] + C.